Engineering Economics The principles Engineering Economics The principles

: Engineering Economics – The principles

Engineering Economics – The principles

F = P ( 1+ i )n P = F 1(1+i )n

Amount End of year borrowed (ID) Interest Amount outlet paid ( ID) (( ID) 0 1000 -- y 1 1 -- 1000*6/100=60 1060 0 y 2 2 -- 1000*6/100=60 1120 0 y 3 3 -- 1000*6/100=60 1180

End of year Amount borrowed (ID) Interest 0 1000 -- 1000 y 1 1 -- 60 1060 y 2 2 -- 60 1120 y 3 3 -- 60 1180 Amount owe

End of year Amount borrowed (ID) Interest 0 3703. 7 -- 3703. 7 1 -- 259. 25 3967. 24 2 -- 519. 08 4226. 78 3 -- 778. 16 4486. 32 4 -- 1038. 16 4745. 86 5 -- 1297 5005. 5 = 5000 ﺍﻟﺮﻗﻢ ﻫﺬﺍ ﻳﻜﻮﻥ ﻥ ﻣﻤﻜﻦ ﺗﺮﺍﻛﻤﻲ ﻏﻴﺮ Amount owe

Example: If we borrowed 2000 ID, at the end of the third year, find the amount of the money which you must paid now. Show the table of calculation? End of year 0 1 2 3 4 5 Amount borrowed (ID) 3707. 7 Interest Amount owe Amount paid ($) ($ ) 3707. 7 0 0 3967. 24 -3707. 7* 7/100 -=259. 25 -259. 25 4226. 78 2000 259. 25 4486. 32 -399. 54 6885. 86 -399. 54 7285. 4 5707, 7 =2000 + 3707, 7 = ﺍﻟﺜﺎﻟﺜﻪ ﺍﻟﺴﻨﻪ ﻧﻬﺎﻳﻪ 0 0 0 7280

Amount End of borrowe year d (ID) 0 3707. 7 1 -2 -3 -4 -5 -- Interest -3703. 7* 7/100 =259. 25 519. 08 259. 25 778. 16 259. 25 1038. 16 259. 25 1297 259. 25 Amount owe paid ($) ($ ) 3707. 7 0 3967. 24 0 4226. 78 0 4486. 32 0 4745. 86 0 5005. 5 5000

What is the effective interest rate of nominal interest rate equal to (6%) compound quarterlies, for one year? ﺳﻨﻪ ﺭﺑﻊ ﻛﻞ ﺍﺳﺎﺱ ﻋﻠﻰ ﺍﻟﻔﺎﺋﺪﺓ ﻳﺤﺴﺐ

F = P(1+i )n F = 400 ( 1 + 0. 08 ) 10 =

Present worth : If the formula S = p(1+i)n is rearrange to express P in terms of S , I , and n , so it will be : p = S/(1+i)n This formula established [P] which must be placed for n periods. What amount of money should be saved in a bank knew to receive (5000$) after 6 years With compound interest rate of 5%? ? Sol: p = S/(1+i)n S =5000$ n = 6 years i = 5% p =? ? ? p = 5000 $ /(1+0. 05)6 = 3731. 07

S = p(1+i)n

Example : Find p 0 , and F 8 and find the equivalent uniform series? ?

PT = 7049. 6 + 5904. 69 +229398. 4 = 35894. 13 $

Example: If a payment of 500 $ at the end of each year for 6 year from now invested with 5% interest rate what is the equivalent of these payment after 8 years from now? A = 500 $ n = 6 i= 5% p = ? ? ?

Equivalence: It is often desirable to compare payments or receipts of different sums of money due at different dates , with each. To do so, it is necessary to convert all sums into their equivalent sums at a definite date, taking into account the effect of interest on time. This can be done by finding the present worth of each sum in order to be able to compare between them on equal bases

Example: If a payment of 500 $ at the end of each year for 6 year from now invested with 5% interest rate what is the equivalent of these payment after 8 years from now? A = 500 $ n = 6 i= 5% p = ? ? ?

Example: An investment of 10, 000 $ ( Buy Machine ) can be made that will produce uniform annual revenue of 5, 310$ for 5 years and then have a positive salvage value of 2, 000$ at the end of 5 years, annual disbursements) will be 3, 000$ at the end of each year for operating and maintaining the project. Draw a cash flow diagram and determine the cumulative cash flow over the 5 -year life of the project? Solution: → cash ﺍﻟﺴﻨﻮﻳﺔ ﺍﻟﻤﺼﺮﻭﻓﺎﺕ {initial investment + annual disbursement outflow → Cash inflow ﻣﺘﺒﻘﻴﺔ ﻗﻴﻤﺔ Annual revenues + Salvage value

5310 -3000 = 2310 ﺍﺳﺘﻠﻢ ﺳﻨﻮﻳﺎ 5310 ﻭﻟﻜﻦ ﻫﻨﺎﻟﻚ ﺻﺮﻓﻴﺎﺕ 3000 ﺍﺫﻥ ﺍﻟﺮﺑﺢ ﺍﻟﺴﻨﻮﻱ ﻳﻜﻮﻥ ﻓﻘﻂ 2310 Accumulative cash flow - 1000 -7690 -5380 -3070 -760 + 3550 Net cash flow -1000 +2310 4310 End of year 0 1 2 3 4 5

Example : IF the maintenance cost of bulldozer amount to ( 200 $ )by the end of the first year of its service ( 250$) by the end of second year ( 300 $), 350, 400 by the end of 3 rd , 4 th and 5 th year respectively with interest rate of 5%, find the equivalent uniform series over a period 5 years ? ?

Their for the uniform series equivalent annual cost of maintenance =200 + 95. 12 = 295. 12 $

Example : For cash Flow diagram shown, calculated the equivalent present worth value (p) with an interest rate = 5%?

3 - To find the future compound amount : F=(G/i) [{(1+i)n -1}/i] -(n. G/i) = (1000/0. 15) [{(1+0. 15)4 -1}/0. 15] -(4*1000/0. 15) = 6622. 22 $

End of year 1 2 3 4 Payment $ 5000 6000 7000 8000 Note ﺍﻟﺪﻓﻌﺎﺕ ﻫﺬﻩ ﻧﺤﻮﻝ ﺍﻟﻰ Annuities

Another Solution

1 --- Find the present Worth for i =7% Base amount is equal to largest amount

1 --- Find the present Worth for i =7% The gradient has value

: ﺍﻣﺎ ﺗﺤﺴﺐ 100, 2000, 500 ≠ G, , , 400 = A Total : 1208

L Useful life the structure in years C The original cost for the structure d The annual cost of depreciation Cn ( Book value )the value at the end of the ( n) year CL the value at the end of the life of structure Dn Depreciation up to age ( n) year n Number of years

End of year 0 1 2 3 4 5 6 7 Depreciation in $ 0 1000 1000 Book value in $ 10000 9000 8000 7000 6000 5000 4000 3000

End of year Declining balance ���� in $ K ���� = �� �� − 1 ∗�� 0 0 0 1 0. 20*10000=2000 2 0. 20*8000=1600 etc. 7 0. 2 524 8 0. 2 420 ���� In $ ���� = �� (1−�� )�� 10000 8000 6400 etc. 2097 1677 ={ 10000(10. 2)^8}=

3. The Sum of the Years' Digits Method : ( SOY) Also caged some times : Sum of the Integers Method

Example : Find the book values and the depreciation allowances at each year, for an equipment that have an initial cost of 10000$ and an estimated salvage value of 2000$ at the end of a useful life of eight years? Annual factor : 2(�� −�� +1) /�� (�� +1) = ----- reverse years / 36 = 8 /36( column 3) ���� = (�� −���� ) ∗ 2(�� −�� +1) /�� (�� +1) =(�� −���� ) * Annual factor ( column 4 2(�� −�� +1) = 2(1 -1+2)=2*1=2>>>> 2/�� (�� +1) = 36 = 2 / 8(8 +1)= 2/72=1/36 ﺍﻻﻭﻟﻰ ﻟﻠﺴﻨﻪ 2(�� −�� +1) = 2(8 -1+1)=2*8=16 16/�� (�� +1) = 36 = 16/72= 2

Part three comparison between Alternatives : ﺍﻻﻗﺘﺼﺎﺩﻳﺔ ﺑﺎﻟﺪﺭﺍﺳﺎﺕ ﻟﻠﻘﻴﺎﻡ ﺍﻟﺮﺋﻴﺴﻴﺔ ﺍﻟﻄﺮﻕ A- Comparison between alternatives with equal lives: 1 - By IRR: Internal rate of return (IRR) 2 - Present Worth Method (PW) 3– Annual worth method: A. W. 4 - The external rate of return method (E. R. R)

Compression between alternatives with unequal lives: First Method: Make the alternatives by equal lives. Second Method: Sinking Method.